(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let a,b be elements of a group G. show that if ab has finite order, then ba has finite order.

2. Relevant equations

3. The attempt at a solution

provided proof:

Let n be the order of ab so that (ab)^{n }= e. Multiplying this equation on the left by b and on the right by a, we find that (ba)^{n+1}= bea = (ba)e. Cancellation of the first factor ba from both sides shows that (ba)^{n}= e, so the order of ba is < n. If the order of ba were less than n, a symmetric argument would show that the order of ab is less than n, contrary to our choice of n. Thus ba has order n also.

okay. so, this is from a chapter on cyclic groups, so i'm assuming that it has to do with, duh, cyclic groups. i know cyclic groups are abelian, and if that were to be assumed from the beginning, i believe this problem would be relatively easy. although, actually, it probably wouldn't even be a problem, since it would be true from the beginning (since we could just say ab=ba, and ab^{n}=ba^{n}

so:

Let n be the order of ab so that (ab)^{n}= e.

got that, that makes sense, i believe that's the definition of order of an element.

WAIT. here's the crux, something i (maybe) just realized as i typed this sentence. since we define the order of ab such that ab^{n}=e, does that imply that G is cyclic, and thus abelian? that would make things easier, if i'm right. so far, a quick internet search has failed me on an answer.

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# Homework Help: Have a proof re. cyclic groups, need a little explaining

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